`1.` Tìm x biết:
$\dfrac{x – 1}{2020}$ `+` $\dfrac{x – 2}{2019}$ `-` $\dfrac{x – 3}{2018}$ `=` $\dfrac{x – 4}{2017}$
`2.` Tìm ` x ∈ Z` biết:
`a)` $\dfrac{1}{1.3}$ `+` $\dfrac{1}{3.5}$ `+` $\dfrac{1}{5.7}$ `+` $\dfrac{1}{( 2x – 1). ( 2x +1)}$ `=` $\dfrac{49}{99}$
`b)` `1 – 3 + 3^{2} – 3^{3} +…+ ( -3)^x =` $\dfrac{9^{1000} – 1}{4}$
Help mì~~
Đáp án:
Bạn xem lại bài 2b nhé phải là `1-3+3^2-3^3+….+(-3)^x=(9^1000+1)/4`
Giải thích các bước giải:
`1,(x-1)/2020+(x-2)/2019-(x-3)/2018=(x-4)/2017`
`<=>(x-1)/2020-1+(x-2)/2019-1=(x-3)/2018-1+(x-4)/2017-`
`<=>(x-2021)/2020+(x-2021)/2019=(x-2021)/2018+(x-2021)/2017`
`<=>(x-2021)(1/2020+1/2019-1/2018-1/2017)=0`
`<=>x=2021` do `1/2020+1/2019-1/2018-1/2017<0`
Vậy `x=2021`
`2a,1/(1.3)+1/(3.5)+1/(5.7)+…..+1/((2x-1)(2x+1))=49/99`
`<=>2/(1.3)+2/(3.5)+2/(5.7)+…..+2/((2x-1)(2x+1))=98/99`
`<=>1-1/3+1/3-1/5+1/5-1/7+….+1/(2x-1)-1/(2x+1)=98/99`
`<=>1-1/(2x+1)=98/99`
`<=>(2x)/(2x+1)=98/99`
`<=>198x=196x+98`
`<=>2x=98`
`<=>x=49`
Vậy `x=49`
`b,Đặt \ A=1-3+3^2-3^3+….+(-3)^x`
`=>3A=3-3^2+3^3-3^4+…..+(-3)^{x+1}`
`=>3A+A=1+(-3)^{x+1}`
`=>4A=1+(-3)^{x+1}`
`=>A=(1+(-3)^{x+1})/4`
Mà `A=(9^1000-1)/4`
`=>1+(-3)^{x+1}=9^1000-1`
`=>(-3)^{x+1}=9^1000-2`
`=>` Không có giá trị nào của x thỏa mãn đề bài.
$\dfrac{x-1}{2020}+\dfrac{x-2}{2019}-\dfrac{x-3}{2018}-\dfrac{x-4}{2017}=0$
$\Leftrightarrow \dfrac{x-2021+2020}{2020}\,\,+\,\,\dfrac{x-2021+2019}{2019}\,\,-\,\,\dfrac{x-2021+2018}{2018}\,\,-\,\,\dfrac{x-2021+2017}{2017}=0$
$\Leftrightarrow \dfrac{x-2021}{2020}\,\,+\,\,1\,\,+\,\,\dfrac{x-2021}{2019}\,\,+\,\,1\,\,-\,\,\dfrac{x-2021}{2018}\,\,-\,\,1\,\,-\,\,\dfrac{x-2021}{2017}\,\,-\,\,1=0$
$\Leftrightarrow \dfrac{x-2021}{2020}\,\,+\,\,\dfrac{x-2021}{2019}\,\,-\,\,\dfrac{x-2021}{2018}\,\,-\,\,\dfrac{x-2021}{2017}=0$
$\Leftrightarrow \left( x-2021 \right)\left( \dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017} \right)=0$
$\Leftrightarrow x-2021=0$ ( Vì $\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{2}{2017}\ne 0$ )
$\Leftrightarrow x=2021$
Bài 2:
a)$\dfrac{1}{1.3}\,+\,\dfrac{1}{3.5}\,+\,\dfrac{1}{5.7}\,+\,…+\dfrac{1}{\left( 2x-1 \right)\left( 2x+1 \right)}=\dfrac{49}{99}$
$\Leftrightarrow \dfrac{1}{2}\left( \dfrac{2}{1.3}\,+\,\dfrac{2}{3.5}\,+\,\dfrac{2}{5.7}\,+\,\dfrac{2}{\left( 2x-1 \right)\left( 2x+1 \right)} \right)=\dfrac{49}{99}$
$\Leftrightarrow \dfrac{1}{1}\,\,\,-\,\,\dfrac{1}{3}\,\,+\,\,\dfrac{1}{3}\,\,-\,\,\dfrac{1}{5}\,\,+\,\,\dfrac{1}{5}\,\,-\,\,\dfrac{1}{7}\,+…\,+\,\,\dfrac{1}{2x-1}\,\,\,\,-\dfrac{1}{2x+1}\,\,=\dfrac{98}{99}$
$\Leftrightarrow \dfrac{1}{1}-\dfrac{1}{2x+1}=\dfrac{98}{99}$
$\Leftrightarrow \dfrac{2x}{2x+1}=\dfrac{98}{99}$
$\Leftrightarrow 99\left( 2x \right)=98\left( 2x+1 \right)$
$\Leftrightarrow 2x=98$
$\Leftrightarrow x=49$
Đặt:
$\,\,\,S=1\,\,-\,\,3\,\,+\,\,{{3}^{2}}\,\,-\,\,{{3}^{3}}\,\,+…\,\,-\,\,{{3}^{x}}$
$3S=\,3\,\,-\,\,{{3}^{2}}\,\,+\,\,{{3}^{3}}\,\,-\,\,{{3}^{4}}\,\,+…\,\,-{{3}^{x+1}}$
$\to S+3S=\left( 1-3+{{3}^{2}}-{{3}^{3}}+…-{{3}^{x}} \right)+\left( 3-{{3}^{2}}+{{3}^{3}}-{{3}^{4}}+…-{{3}^{x+1}} \right)$
$\to 4S=1-{{3}^{x+1}}$
$\to S=\dfrac{1-{{3}^{x+1}}}{4}$
$\to \dfrac{{{9}^{1000}}-1}{4}=\dfrac{1-{{3}^{x+1}}}{4}$
$\to {{9}^{1000}}-1=1-{{3}^{x+1}}$
$\to {{3}^{x+1}}=2-{{9}^{1000}}$
${{3}^{x+1}}$ chia hết cho $3$
$2$ không chia hết cho $3$
${{9}^{1000}}$ chia hết cho $3$
Nên không có giá trị $x$
Đổi đề một chút:
$S=\dfrac{1-{{3}^{x+1}}}{4}=\dfrac{1-{{9}^{1000}}}{4}$
$\to 1-{{3}^{x+1}}=1-{{9}^{1000}}$
$\to {{3}^{x+1}}={{9}^{1000}}$
$\to {{3}^{x+1}}={{3}^{2000}}$
$\to x=1999$